If $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \\ 2x + 5, & 3 < x < 4 \end{cases}$,the equation whose roots are $\lim_{x \to 3^-} f(x)$ and $\lim_{x \to 3^+} f(x)$ is

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ where $f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 0, & x=0 \end{cases}$

$\lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}}$ is equal to:

Let $f(x) = \frac{\ln(x^2 + e^x)}{\ln(x^4 + e^{2x})}$. If $\lim_{x \to \infty} f(x) = l$ and $\lim_{x \to -\infty} f(x) = m$,then:

Let $[.]$ denote the greatest integer function. Assertion $(A) : \lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$. Reason $(R) : f(x) = x - 1, g(x) = [x], h(x) = x$ and $\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{h(x)}{x} = 1$.

If $g(x) = \frac{x}{[x]}$ for $x > 2$,then $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ is equal to